Given an invertible real matrix $A$ and real column vectors $b$ and $c$.

For which $A$,$b$ and $c$ are all corresponding principal minors of $B = A-bc^T$ and $A^{-1}$ equal?

According to a result by Loewy, this is true if $B$ and $A^{-1}$ are diagonally similar with transpose (plus some extra conditions). We can assume that both $A$ and $A^{-1}$ are adjacency matrices of fully connected graphs, i.e., all entries are non-zero.

My main interests are:

1. For which matrices $A$ is the problem solvable?
2. Given a matrix $A$, how obtain $b$ and $c$ ?
3. As a general characterization of $A$ might be difficult, I am particularly interested in a solution of the form $A = G O G$, with diagonal matrix $G \neq I$ and orthogonal $O$. Can you think of a class of matrices $O$ which simplifies this problem?

Motivation: The problem arises in control theory, where transfer function from a state-space formulation is:

$$ H(z) = \frac{\det(A) \det(D(z) - (A - bc^T))}{\det(D(z) - A)}, $$ where $D(z) = diag([z^{m_1},\dots,z^{m_n}])$ for integer $m_i$. The goal is now to choose $A$, $b$ and $c$ such that $|H(z)|=1$ for all $z$. This is true if the numerator and denominator of $H(z)$ are "flipped", i.e.,

$$ flip(\det(D(z) - A)) = \det(A) \det(D(z) - A^{-1}). $$

Thus, for any $m_i$, we need: $$ \det(D(z) - A^{-1}) = \det(D(z) - (A - bc^T)), $$ which is true if all principal minors of $B$ and $A^{-1}$ are equivalent.

Attempt Following the work of Engel and Schneider and assuming fully connectedness of $A$: Let $H = A \div B$, where $\div$ is element-wise. In Corrolary 3.11., it can be easily seen (from the fully connectedness) that $H$ is diagonally similar to a matrix of only 1s. Thus, for $A$ and $B$ to be diagonally similar, we need $H_{ii} = 1$ for all $i$. Hence, $$ c_i = \frac{ A_{ii} - (A^{-1})_{ii} }{b_{i}}. $$ Remains to determine $b$.

Example Size=2

For

$$ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} $$ and $b = [3, 4.5]^T$ and $c = [1,1]^T$. Then

$$ B = \begin{bmatrix} -2 & -1 \\ -1.5 & -0.5 \end{bmatrix} $$ which is diagonally similar to $$ A^{-1} = \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix} $$